- #1
Sennap
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Homework Statement
(Translated, sorry my English) Find how fast does the radius of a circle get smaller when its area is 75*pi cm^2 and it (the area) gets smaller at a rate of 2*pi cm^2/s?
Homework Equations
Not really any given equations, but I used these:
A = r^2 * pi
s = v*t + s0 (If v is a constant)
t=A/Va (where t is the time when A=0)
The Attempt at a Solution
From now on I'll call area, A = 75*pi cm^2 and the rate v_a = 2*pi cm^2/s.
I gave myself that the speed (v) was constant, that shouldn't really change the answer.
So s = dv/dt = v*t + s0
By using A = r^2 * pi we know that:
r^2 = A/pi
r = sqrt(A/pi) and since at t = 0 then A = 75*pi so
r = sqrt(75) at t = 0, let's call r from now on s.
Let's find the time until A = 0, and therefore s=0
t = A/v_a = 75*pi/(2*pi) = 37,5 sec
So in 37,5 sec the circle area will be zero with is radius. Therefore we can see that:
s = v*t + s0
0 = v*37,5 + sqrt(75)
-sqrt(75) = v*37,5
v = -sqrt(75)/37,5
v = -2*sqrt(75)/75
v = -2/sqrt(75)
So my final answer was v = -2/sqrt(75) (or v = -2/(5*sqrt(3))But according to my teacher the correct answer is v = -1/sqrt(75).--
Here's how my teacher did it:
We know that F(t) = pi * r(t)^2 and
(1) F'(t0) = 2*pi*r(t0)*r'(t0) = -2*pi
(2) F(t0) = pi*r(t)^2 = 75*pi
for a specific t0.
According to (2) we know that r(t0) = sqrt(75) = 5*sqrt(3) and according to (1) we get r'(t0) = -1/r(t0) = -1/sqrt(75)
--
He called my solution "a waste of time" although he said it should work too, but couldn't explain to me why I didn't get the same answer.
Where did I go wrong (or even better, where did he go wrong?)